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Always be mindful of the kindness and not the faults of others.

Always be mindful of the kindness and not the faults of others. One-way Anova: Inferences about More than Two Population Means. Model and test for one-way anova Assumption checking Nonparamateric alternative. Analysis of Variance & One Factor Designs (One-Way ANOVA).

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Always be mindful of the kindness and not the faults of others.

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  1. Always be mindful of the kindness and not the faults of others.

  2. One-way Anova: Inferences about More than Two Population Means Model and test for one-way anova Assumption checking Nonparamateric alternative

  3. Analysis of Variance & One Factor Designs (One-Way ANOVA) Y= RESPONSE VARIABLE (of numerical type) (e.g. battery lifetime) X = EXPLANATORY VARIABLE (of categorical type) (A possibly influential FACTOR) (e.g. brand of battery) OBJECTIVE: To determine the impact of X on Y

  4. Completely Randomized Design (CRD) • Goal: to study the effect of Factor X • The same # of observations are taken randomly and independently from the individuals at each level of Factor X • i.e. n1=n2=…nc (c levels)

  5. 1 2 3 4 5 6 7 8 1.8 4.2 8.6 7.0 4.2 4.2 7.8 9.0 5.0 5.4 4.6 5.0 7.8 4.2 7.0 7.4 1.0 4.2 4.2 9.0 6.6 5.4 9.8 5.8 5.8 2.6 4.6 5.8 7.0 6.2 4.6 8.2 7.4 Example: Y = LIFETIME (HOURS) BRAND 3 replications per level

  6. Analysis of Variance

  7. StatisticalModel C “levels” OF BRAND R observations for each level 1 2 • • •  •  •  • • • R 1 2 • • • • C Y11 Y12 • • • • • • •Y1R Yij = + i + ij i = 1, . . . . . , C j = 1, . . . . . , R Y21 • • • • • • YcI • • • • • Yij YcR •   •  •   •    •   •    •    • 

  8. Where = OVERALL AVERAGE i = index for FACTOR (Brand) LEVEL j= index for “replication” i = Differential effect associated with ith level of X (Brand i) = mi – m and ij = “noise” or “error” due to other factors associated with the (i,j)th data value. mi = AVERAGE associated with ith level of X (brand i) m= AVERAGE of mi ’s.

  9. Yij =  + i + ij By definition, i = 0 C i=1 The experiment produces R x C Yij data values. The analysis produces estimates of ,c. (We can then get estimates of the ij by subtraction).

  10. Let Y1, Y2, etc., be level means Y • = Y i /C = “GRAND MEAN” (assuming same # data points in each column) (otherwise, Y • = mean of all the data) c i=1

  11. MODEL: Yij =  + i + ij Y• estimates  Yi - Y • estimatesi (= mi – m) (for all i) These estimates are based on Gauss’ (1796) PRINCIPLE OF LEAST SQUARES and on COMMON SENSE

  12. MODEL: Yij =  + j + ij If you insert the estimates into the MODEL, (1) Yij = Y • + (Yj - Y • ) + ij. < it follows that our estimate of ij is (2) ij = Yij – Yj, called residual <

  13. Then, Yij = Y• + (Yi - Y• ) + ( Yij - Yi) or, (Yij - Y• ) = (Yi - Y•) + (Yij - Yi ) { { { (3) Variability in Y associated with all other factors Variability in Y associated with X TOTAL VARIABILITY in Y + =

  14. If you square both sides of (3), and double sum both sides (over i and j), you get, [after some unpleasant algebra, but lots of terms which “cancel”] {{ C C R C R (Yij - Y• )2 = R •  (Yi - Y•)2 + (Yij - Yi)2 { i=1 i=1 j=1 i=1 j=1 ( SSW (SSE) SUM OF SQUARES WITHIN SAMPLES TSS TOTAL SUM OF SQUARES SSB SUM OF SQUARES BETWEEN SAMPLES + + = = ( ( ( ( (

  15. ANOVA TABLE SOURCE OF VARIABILITY Mean square (M.S.) SSQ DF Between samples (due to brand) SSB SSB C - 1 = MSB C - 1 Within samples (due to error) SSW MSW = (R - 1) • C SSW (R-1)•C TOTAL TSS RC -1

  16. 1 2 3 4 5 6 7 8 1.8 4.2 8.6 7.0 4.2 4.2 7.8 9.0 5.0 5.4 4.6 5.0 7.8 4.2 7.0 7.4 1.0 4.2 4.2 9.0 6.6 5.4 9.8 5.8 5.8 2.6 4.6 5.8 7.0 6.2 4.6 8.2 7.4 Example: Y = LIFETIME (HOURS) BRAND 3 replications per level SSB= 3 ( [2.6 - 5.8]2 + [4.6 - 5.8] 2+ • • • + [7.4 - 5.8]2) = 3 (23.04) = 69.12

  17. SSW =? (1.8 - 2.6)2 = .64 (4.2 - 4.6)2 =.16 (9.0 -7.4)2 = 2.56 (5.0 - 2.6)2 = 5.76 (5.4 - 4.6)2= .64 • • • • (7.4 - 7.4)2 = 0 (1.0 - 2.6)2 = 2.56 (4.2 - 4.6)2= .16 (5.8 - 7.4)2 = 2.56 8.96 .96 5.12 Total of (8.96 + .96 + • • • + 5.12), SSW = 46.72

  18. ANOVA TABLE Source of Variability df M.S. SSQ 7 = 8 - 1 69.12 BRAND 9.87 ERROR 2.92 16 = 2 (8) 46.72 TOTAL 115.84 23 = (3 • 8) -1

  19. We can show: “VCOL” { E (MSB) = 2+ MEASURE OF DIFFERENCES AMONG LEVEL MEANS ( R ( • (i - )2 { C-1 i E (MSW) = 2 (Assuming Yij follows N(j , 2) and they are independent)

  20. E ( MSBC ) = 2 +VCOL E ( MSW) = 2 This suggests that There’s some evidence of non-zero VCOL, or “level of X affects Y” if MSBC > 1 , MSW if MSBC No evidence that VCOL > 0, or that “level of X affects Y” < 1 , MSW

  21. With HO: Level of X has no impact on Y HI: Level of X does have impact on Y, We need MSBC > > 1 MSW to reject HO.

  22. More Formally, HO: 1 = 2 = • • • c = 0 HI: not all j = 0 OR (All level means are equal) HO: 1 = 2 = • • • • c HI: not all j are EQUAL

  23. The distribution of MSB = “Fcalc” , is MSW The F - distribution with (C-1, (R-1)C) degrees of freedom  Assuming HO true. C = Table Value

  24. In our problem: ANOVA TABLE Source of Variability M.S. Fcalc SSQ df 7 69.12 BRAND 9.87 3.38 ERROR 2.92 = 9.87 2.92 16 46.72

  25. F table: table 8 = .05 C = 2.66 3.38 (7,16 DF)

  26. Hence, at  = .05, Reject Ho . (i.e., Conclude that level of BRAND does have an impact on battery lifetime.)

  27. MINITAB INPUT life brand 1.8 1 5.0 1 1.0 1 4.2 2 5.4 2 4.2 2 . . . . . . 9.0 8 7.4 8 5.8 8

  28. ONE FACTOR ANOVA (MINITAB) MINITAB: STAT>>ANOVA>>ONE-WAY Analysis of Variance for life Source DF SS MS F P brand 7 69.12 9.87 3.38 0.021 Error 16 46.72 2.92 Total 23 115.84 Estimate of the common variances^2

  29. Assumptions MODEL: Yij = + i + ij 1.) the ij are indep. random variables 2.) Each ij is Normally Distributed E(ij) = 0 for all i, j 3.) 2(ij) = constant for all i, j Run order plot Normality plot & test Residual plot & test

  30. Diagnosis: Normality • The points on the normality plot must more or less follow a line to claim “normal distributed”. • There are statistic tests to verify it scientifically. • The ANOVA method we learn here is not sensitive to the normality assumption. That is, a mild departure from the normal distribution will not change our conclusions much. Normal probability plot & normality test of residuals

  31. Minitab: stat>>basic statistics>>normality test

  32. Diagnosis: Constant Variances • The points on the residual plot must be more or less within a horizontal band to claim “constant variances”. • There are statistic tests to verify it scientifically. • The ANOVA method we learn here is not sensitive to the constant variances assumption. That is, slightly different variances within groups will not change our conclusions much. Tests and Residual plot: fitted values vs. residuals

  33. Minitab: Stat >> Anova >> One-way

  34. Minitab: Stat>> Anova>> Test for Equal variances

  35. Diagnosis: Randomness/Independence • The run order plot must show no “systematic” patterns to claim “randomness”. • There are statistic tests to verify it scientifically. • The ANOVA method is sensitive to the randomness assumption. That is, a little level of dependence between data points will change our conclusions a lot. Run order plot: order vs. residuals

  36. Minitab: Stat >> Anova >> One-way

  37. KRUSKAL - WALLIS TEST (Non - Parametric Alternative) HO: The probability distributions are identical for each level of the factor HI: Not all the distributions are the same

  38. Brand ABC 32 32 28 30 32 21 30 26 15 29 26 15 26 22 14 23 20 14 20 19 14 19 16 11 18 14 9 12 14 8 BATTERY LIFETIME (hours) (each column rank ordered, for simplicity) Mean: 23.9 22.1 14.9 (here, irrelevant!!)

  39. HO: no difference in distribution among the three brands with respect to battery lifetime HI: At least one of the 3 brands differs in distribution from the others with respect to lifetime

  40. Ranks in ( ) Brand ABC 32 (29) 32 (29) 28 (24) 30 (26.5) 32 (29) 21 (18) 30 (26.5) 26 (22) 15 (10.5) 29 (25) 26 (22) 15 (10.5) 26 (22) 22 (19) 14 (7) 23 (20) 20 (16.5) 14 (7) 20 (16.5) 19 (14.5) 14 (7) 19 (14.5) 16 (12) 11 (3) 18 (13) 14 (7) 9 (2) 12 (4) 14 (7) 8 (1) T1 = 197T2 = 178 T3 = 90 n1 = 10 n2 = 10 n3 = 10

  41. TEST STATISTIC: K 12 •  (Tj2/nj ) - 3 (N + 1) H = N (N + 1) j = 1 nj = # data values in column j N = nj K = # Columns (levels) Tj = SUM OF RANKS OF DATA ON COL j When all DATA COMBINED (There is a slight adjustment in the formula as a function of the number of ties in rank.) K j = 1

  42. H = [ 12 197 2 178 2 902 30 (31) 10 10 10 [ + + - 3 (31) = 8.41 (with adjustment for ties, we get 8.46)

  43. c21-adf  = .05 = F1-adf, df 5.99 8.41 = H What do we do with H? We can show that, under HO , H is well approximated by a 2 distribution with df = K - 1. Here, df = 2, and at = .05, the critical value = 5.99 8 Reject HO; conclude that mean lifetime NOT the same for all 3 BRANDS

  44. Minitab: Stat >> Nonparametrics >> Kruskal-Wallis • Kruskal-Wallis Test: life versus brand • Kruskal-Wallis Test on life • brand N Median AveRank Z • 1 3 1.800 4.5 -2.09 • 2 3 4.200 7.8 -1.22 • 3 3 4.600 11.8 -0.17 • 4 3 7.000 16.5 1.05 • 5 3 6.600 13.3 0.22 • 6 3 4.200 7.8 -1.22 • 7 3 7.800 20.0 1.96 • 8 3 7.400 18.2 1.48 • Overall 24 12.5 • H = 12.78 DF = 7 P = 0.078 • H = 13.01 DF = 7 P = 0.072 (adjusted for ties)

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